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Math Problems
Grade 6
Area of quadrilaterals and triangles: word problems
The surface area of a cube is increasing at a rate of
15
15
15
square meters per hour.
\newline
At a certain instant, the surface area is
24
24
24
square meters.
\newline
What is the rate of change of the volume of the cube at that instant (in cubic meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
15
2
\frac{15}{2}
2
15
\newline
(B)
(
15
)
3
(\sqrt{15})^{3}
(
15
)
3
\newline
(C)
5
8
\frac{5}{8}
8
5
\newline
(D)
8
8
8
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The volume of a cube is increasing at a rate of
18
18
18
cubic meters per hour.
\newline
At a certain instant, the volume is
8
8
8
cubic meters.
\newline
What is the rate of change of the surface area of the cube at that instant (in square meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
3
2
\frac{3}{2}
2
3
\newline
(B)
24
24
24
\newline
(C)
36
36
36
\newline
(D)
(
18
3
)
2
(\sqrt[3]{18})^{2}
(
3
18
)
2
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The circumference of a circle is increasing at a rate of
π
2
\frac{\pi}{2}
2
π
meters per hour.
\newline
At a certain instant, the circumference is
12
π
12 \pi
12
π
meters.
\newline
What is the rate of change of the area of the circle at that instant (in square meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
π
4
\frac{\pi}{4}
4
π
\newline
(B)
36
π
36 \pi
36
π
\newline
(C)
3
π
3 \pi
3
π
\newline
(D)
6
6
6
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The radius of the base of a cylinder is increasing at a rate of
1
1
1
meter per hour and the height of the cylinder is decreasing at a rate of
4
4
4
meters per hour.
\newline
At a certain instant, the base radius is
5
5
5
meters and the height is
8
8
8
meters.
\newline
What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
180
π
180 \pi
180
π
\newline
(B)
20
π
20 \pi
20
π
\newline
(C)
−
20
π
-20 \pi
−
20
π
\newline
(D)
−
180
π
-180 \pi
−
180
π
\newline
The volume of a cylinder with base radius
r
r
r
and height
h
h
h
is
π
r
2
h
\pi r^{2} h
π
r
2
h
.
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One diagonal of a rhombus is decreasing at a rate of
7
7
7
centimeters per minute and the other diagonal of the rhombus is increasing at a rate of
10
10
10
centimeters per minute.
\newline
At a certain instant, the decreasing diagonal is
4
4
4
centimeters and the increasing diagonal is
6
6
6
centimeters.
\newline
What is the rate of change of the area of the rhombus at that instant (in square centimeters per minute)?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
1
-1
−
1
\newline
(B)
16
\mathbf{1 6}
16
\newline
(C)
−
16
-16
−
16
\newline
(D)
1
1
1
\newline
The area of a rhombus with diagonals
d
1
d_{1}
d
1
and
d
2
d_{2}
d
2
is
d
1
d
2
2
\frac{d_{1} d_{2}}{2}
2
d
1
d
2
.
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The area of a square is increasing at a rate of
20
20
20
square meters per hour.
\newline
At a certain instant, the area is
49
49
49
square meters.
\newline
What is the rate of change of the perimeter of the square at that instant (in meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
40
7
\frac{40}{7}
7
40
\newline
(B)
2
5
2 \sqrt{5}
2
5
\newline
(C)
28
28
28
\newline
(D)
7
7
7
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The side length of a square is increasing at a rate of
15
15
15
millimeters per second.
\newline
At a certain instant, the side length is
22
22
22
millimeters.
\newline
What is the rate of change of the area of the square at that instant (in square millimeters per second)?
\newline
Choose
1
1
1
answer:
\newline
(A)
660
660
660
\newline
(B)
484
484
484
\newline
(C)
30
30
30
\newline
(D)
225
225
225
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The radius of a circle is decreasing at a rate of
6
6
6
.
5
5
5
meters per minute.
\newline
At a certain instant, the radius is
12
12
12
meters.
\newline
What is the rate of change of the area of the circle at that instant (in square meters per minute)?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
156
π
-156 \pi
−
156
π
\newline
(B)
−
144
π
-144 \pi
−
144
π
\newline
(C)
−
288
π
-288 \pi
−
288
π
\newline
(D)
−
42.25
π
-42.25 \pi
−
42.25
π
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The radius of a circle is increasing at a rate of
3
3
3
centimeters per second.
\newline
At a certain instant, the radius is
8
8
8
centimeters.
\newline
What is the rate of change of the area of the circle at that instant (in square centimeters per second)?
\newline
Choose
1
1
1
answer:
\newline
(A)
48
π
48 \pi
48
π
\newline
(B)
9
π
9 \pi
9
π
\newline
(C)
64
π
64 \pi
64
π
\newline
(D)
192
π
192 \pi
192
π
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The radius of a sphere is increasing at a rate of
7
7
7
.
5
5
5
meters per minute.
\newline
At a certain instant, the radius is
5
5
5
meters.
\newline
What is the rate of change of the surface area of the sphere at that instant (in square meters per minute)?
\newline
Choose
1
1
1
answer:
\newline
(A)
330
π
330 \pi
330
π
\newline
(B)
100
π
100 \pi
100
π
\newline
(C)
300
π
300 \pi
300
π
\newline
(D)
225
π
225 \pi
225
π
\newline
The surface area of a sphere with radius
r
r
r
is
4
π
r
2
4 \pi r^{2}
4
π
r
2
.
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The area of a circle is increasing at a rate of
8
π
8 \pi
8
π
square meters per hour.
\newline
At a certain instant, the area is
36
π
36 \pi
36
π
square meters.
\newline
What is the rate of change of the circumference of the circle at that instant (in meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
12
π
12 \pi
12
π
\newline
(B)
π
6
\frac{\pi}{6}
6
π
\newline
(C)
4
2
4 \sqrt{2}
4
2
\newline
(D)
4
π
3
\frac{4 \pi}{3}
3
4
π
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The radius of the base of a cylinder is increasing at a rate of
1
1
1
meter per hour and the height of the cylinder is decreasing at a rate of
4
4
4
meters per hour.
\newline
At a certain instant, the base radius is
5
5
5
meters and the height is
8
8
8
meters.
\newline
What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
20
π
-20 \pi
−
20
π
\newline
(B)
−
180
π
-180 \pi
−
180
π
\newline
(C)
20
π
20 \pi
20
π
\newline
(D)
180
π
180 \pi
180
π
\newline
The volume of a cylinder with base radius
r
r
r
and height
h
h
h
is
π
r
2
h
\pi r^{2} h
π
r
2
h
.
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The surface area of a sphere is increasing at a rate of
14
π
14 \pi
14
π
square meters per hour.
\newline
At a certain instant, the surface area is
36
π
36 \pi
36
π
square meters.
\newline
What is the rate of change of the volume of the sphere at that instant (in cubic meters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
36
π
36 \pi
36
π
\newline
(B)
(
14
π
)
3
(\sqrt{14 \pi})^{3}
(
14
π
)
3
\newline
(C)
21
π
21 \pi
21
π
\newline
(D)
7
12
\frac{7}{12}
12
7
\newline
The surface area of a sphere with radius
r
r
r
is
4
π
r
2
4 \pi r^{2}
4
π
r
2
.
\newline
The volume of a sphere with radius
r
r
r
is
4
3
π
r
3
\frac{4}{3} \pi r^{3}
3
4
π
r
3
.
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The radius of a circle is decreasing at a rate of
6
6
6
.
5
5
5
meters per minute.
\newline
At a certain instant, the radius is
12
12
12
meters.
\newline
What is the rate of change of the area of the circle at that instant (in square meters per minute)?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
288
π
-288 \pi
−
288
π
\newline
(B)
−
156
π
-156 \pi
−
156
π
\newline
(C)
−
42.25
π
-42.25 \pi
−
42.25
π
\newline
(D)
−
144
π
-144 \pi
−
144
π
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The radius of a circle is increasing at a rate of
3
3
3
centimeters per second.
\newline
At a certain instant, the radius is
8
8
8
centimeters.
\newline
What is the rate of change of the area of the circle at that instant (in square centimeters per second)?
\newline
Choose
1
1
1
answer:
\newline
(A)
192
π
192 \pi
192
π
\newline
(B)
64
π
64 \pi
64
π
\newline
(C)
9
π
9 \pi
9
π
\newline
(D)
48
π
48 \pi
48
π
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The radius of a sphere is increasing at a rate of
7
7
7
.
5
5
5
meters per minute.
\newline
At a certain instant, the radius is
5
5
5
meters.
\newline
What is the rate of change of the surface area of the sphere at that instant (in square meters per minute)?
\newline
Choose
1
1
1
answer:
\newline
(A)
100
π
100 \pi
100
π
\newline
(B)
330
π
330 \pi
330
π
\newline
(C)
300
π
300 \pi
300
π
\newline
(D)
225
π
225 \pi
225
π
\newline
The surface area of a sphere with radius
r
r
r
is
4
π
r
2
4 \pi r^{2}
4
π
r
2
.
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The radius of the base of a cylinder is decreasing at a rate of
9
9
9
millimeters per hour and the height of the cylinder is increasing at a rate of
2
2
2
millimeters per hour.
\newline
At a certain instant, the base radius is
8
8
8
millimeters and the height is
3
3
3
millimeters.
\newline
What is the rate of change of the surface area of the cylinder at that instant (in square millimeters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
155
π
155 \pi
155
π
\newline
(B)
310
π
310 \pi
310
π
\newline
(C)
−
155
π
-155 \pi
−
155
π
\newline
(D)
−
310
π
-310 \pi
−
310
π
\newline
The surface of a cylinder with base radius
r
r
r
and height
h
h
h
is
2
π
r
h
+
2
π
r
2
.
2 \pi r h+2 \pi r^{2} \text {. }
2
π
r
h
+
2
π
r
2
.
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The radius of the base of a cylinder is decreasing at a rate of
9
9
9
millimeters per hour and the height of the cylinder is increasing at a rate of
2
2
2
millimeters per hour.
\newline
At a certain instant, the base radius is
8
8
8
millimeters and the height is
3
3
3
millimeters.
\newline
What is the rate of change of the surface area of the cylinder at that instant (in square millimeters per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
310
π
310 \pi
310
π
\newline
(B)
−
310
π
-310 \pi
−
310
π
\newline
(C)
155
π
155 \pi
155
π
\newline
(D)
−
155
π
-155 \pi
−
155
π
\newline
The surface of a cylinder with base radius
r
r
r
and height
h
h
h
is
2
π
r
h
+
2
π
r
2
.
2 \pi r h+2 \pi r^{2} \text {. }
2
π
r
h
+
2
π
r
2
.
Get tutor help
A pendulum is swinging back and forth. After
t
t
t
seconds, the horizontal distance from the bob to the place where it was released is given by
\newline
H
(
t
)
=
7
−
7
cos
(
2
π
(
t
−
2
)
20
)
.
H(t)=7-7 \cos \left(\frac{2 \pi(t-2)}{20}\right) .
H
(
t
)
=
7
−
7
cos
(
20
2
π
(
t
−
2
)
)
.
\newline
How often does the bob cross its midline? Give an exact answer.
\newline
Every
□
\square
□
seconds
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A piece of paper is to display
128
128
128
square inches of text. If there are to be one-inch margins on both sides and two-inch margins at the bottom and top, what are the dimensions of the smallest piece of paper (by area) that can be used?
\newline
Choose
1
1
1
answer:
\newline
(A)
8
′
′
×
16
′
′
8 \prime \prime \times 16 \prime \prime
8′′
×
16′′
\newline
(B)
10
′
′
×
15
′
′
10 \prime \prime \times 15 \prime \prime
10′′
×
15′′
\newline
(C)
10
′
′
×
18
′
′
10 \prime \prime \times 18 \prime \prime
10′′
×
18′′
\newline
(D)
10
′
′
×
20
′
′
10 \prime \prime \times 20 \prime \prime
10′′
×
20′′
\newline
(E) None of these
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M
(
t
)
M(t)
M
(
t
)
models the distance (in millions of
k
m
\mathrm{km}
km
) from Mars to the Sun
t
t
t
days after it's at its furthest point.
\newline
M
(
t
)
=
21
cos
(
2
π
687
t
)
+
228
M(t)=21 \cos \left(\frac{2 \pi}{687} t\right)+228
M
(
t
)
=
21
cos
(
687
2
π
t
)
+
228
\newline
What does the solution set for
\newline
240
=
21
cos
(
2
π
687
t
)
+
228
240=21 \cos \left(\frac{2 \pi}{687} t\right)+228
240
=
21
cos
(
687
2
π
t
)
+
228
\newline
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The farthest distance, in millions of kilometers, that Mars reaches from the sun
\newline
(B) The least number of days after when Mars is at its furthest point when it is
240
240
240
million kilometers from the sun
\newline
(C) The distance, in millions of kilometers, that Mars is from the sun after
240
240
240
days
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In Figure
1
1
1
, each capacitance
C
1
C_{1}
C
1
is
6.0
μ
F
6.0 \mu \mathrm{F}
6.0
μ
F
, and each capacitance
C
2
C_{2}
C
2
is
4
μ
F
4 \mu \mathrm{F}
4
μ
F
.
\newline
(a) Calculate the equivalent capacitance of the network between points
a
a
a
and
b
b
b
.
\newline
(b) Calculate the charge on each of the three capacitors nearest
a
a
a
and
b
b
b
when
V
a
b
=
420
V
V_{a b}=420 \mathrm{~V}
V
ab
=
420
V
.
\newline
(c) With
420
V
420 \mathrm{~V}
420
V
across
a
a
a
and
b
b
b
, compute
6.0
μ
F
6.0 \mu \mathrm{F}
6.0
μ
F
2
2
2
.
\newline
Figure
1
1
1
: for Question
2
2
2
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You want to know how many stories a skyscraper is, knowing the angle to the
5
th
5^{\text{th}}
5
th
floor, and the angle to the top floor.Find thr length of
x
x
x
if the smaller angle is
3
0
∘
30^\circ
3
0
∘
angle and the larger angle is
6
0
∘
60^\circ
6
0
∘
angle
Get tutor help
13
13
13
.
\newline
In the diagram,
P
P
P
is the point of intersection of the curves
y
=
e
2
x
y=\mathrm{e}^{2 x}
y
=
e
2
x
and
y
=
e
2
−
x
y=\mathrm{e}^{2-x}
y
=
e
2
−
x
.
\newline
(i) Find the
x
x
x
-coordinate of
P
P
P
.
\newline
Hence evaluate, correct to two decimal places,
\newline
(ii) the area of the shaded region,
\newline
(iii) the area of the region bounded by the two curves and the
y
y
y
-axis.
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Question
1
1
1
: (
3
3
3
Marks)
\newline
The table below shows the
C
O
2
\mathrm{CO}_{2}
CO
2
emissions (tonnes per capita) for the years
2014
2014
2014
and
2020
2020
2020
for the
28
28
28
countries in the European Union.
\newline
\begin{tabular}{|l|c|c|}
\newline
\hline \multirow{
2
2
2
}{*}{ European Union } & \multicolumn{
2
2
2
}{c|}{ CO
2
2
2
emissions (tonnes per capita) } \\
\newline
\hline Country Name &
2014
2014
2014
&
2020
2020
2020
\\
\newline
\hline Austria &
6
6
6
.
87
87
87
&
6
6
6
.
73
73
73
\\
\newline
\hline Belgium &
8
8
8
.
33
33
33
&
7
7
7
.
23
23
23
\\
\newline
\hline Bulgaria &
5
5
5
.
87
87
87
&
5
5
5
.
39
39
39
\\
\newline
\hline Croatia &
3
3
3
.
97
97
97
&
4
4
4
.
14
14
14
\\
\newline
\hline Cyprus &
5
5
5
.
26
26
26
&
5
5
5
.
38
38
38
\\
\newline
\hline Czech Republic &
9
9
9
.
17
17
17
&
8
8
8
.
22
22
22
\\
\newline
\hline Denmark &
5
5
5
.
94
94
94
&
4
4
4
.
52
52
52
\\
\newline
\hline Estonia &
14
14
14
.
85
85
85
&
7
7
7
.
88
88
88
\\
\newline
\hline Finland &
8
8
8
.
66
66
66
&
7
7
7
.
09
09
09
\\
\newline
\hline France &
5
5
5
.
08
08
08
&
4
4
4
.
24
24
24
\\
\newline
\hline Germany &
8
8
8
.
89
89
89
&
7
7
7
.
69
69
69
\\
\newline
\hline Greece &
6
6
6
.
18
18
18
&
5
5
5
.
01
01
01
\\
\newline
\hline Hungary &
4
4
4
.
27
27
27
&
5
5
5
.
00
00
00
\\
\newline
\hline Ireland &
7
7
7
.
31
31
31
&
6
6
6
.
73
73
73
\\
\newline
\hline Italy &
5
5
5
.
27
27
27
&
5
5
5
.
02
02
02
\\
\newline
\hline Latvia &
3
3
3
.
50
50
50
&
3
3
3
.
59
59
59
\\
\newline
\hline Lithuania &
4
4
4
.
38
38
38
&
5
5
5
.
07
07
07
\\
\newline
\hline Luxembourg &
17
17
17
.
36
36
36
&
13
13
13
.
06
06
06
\\
\newline
\hline Malta &
5
5
5
.
40
40
40
&
3
3
3
.
61
61
61
\\
\newline
\hline Netherlands &
9
9
9
.
92
92
92
&
8
8
8
.
06
06
06
\\
\newline
\hline Poland &
7
7
7
.
52
52
52
&
7
7
7
.
92
92
92
\\
\newline
\hline Portugal &
4
4
4
.
33
33
33
&
3
3
3
.
96
96
96
\\
\newline
\hline Romania &
3
3
3
.
52
52
52
&
3
3
3
.
72
72
72
\\
\newline
\hline Slovak Republic &
5
5
5
.
66
66
66
&
5
5
5
.
63
63
63
\\
\newline
\hline Slovenia &
6
6
6
.
21
21
21
&
6
6
6
.
04
04
04
\\
\newline
\hline Spain &
5
5
5
.
03
03
03
&
4
4
4
.
47
47
47
\\
\newline
\hline Sweden &
4
4
4
.
48
48
48
&
3
3
3
.
83
83
83
\\
\newline
\hline United Kingdom &
6
6
6
.
50
50
50
&
4
4
4
.
85
85
85
\\
\newline
\hline & & \\
\newline
\hline
\newline
\end{tabular}
\newline
(Table
1
1
1
)
\newline
3
3
3
Get tutor help
12
12
12
)
A
B
C
D
A B C D
A
BC
D
is a square. The length of
B
C
B C
BC
is
3
3
3
times the length of
E
B
E B
EB
.
\newline
a) What fraction of square
A
B
C
D
A B C D
A
BC
D
is the shaded part? (Hint: Draw lines to divide
A
B
C
D
A B C D
A
BC
D
into equal parts.)
\newline
b) If
E
B
=
3
c
m
E B=3 \mathrm{~cm}
EB
=
3
cm
, find the area of the unshaded part.
\newline
3
×
3
=
9
9
×
9
=
81
3
×
9
÷
2
=
\begin{array}{l} 3 \times 3=9 \\ 9 \times 9=81 \\ 3 \times 9 \div 2= \end{array}
3
×
3
=
9
9
×
9
=
81
3
×
9
÷
2
=
\newline
Ans: a)
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Estelle B
240
240
240
\newline
Ali |
240
240
240
\newline
You
\newline
240
240
240
\newline
Mookyungs
246
246
246
\newline
Almaa M
247
247
247
\newline
Use the net below to calculate the surface area of this rectangular prism.
Get tutor help
The figure below is a triangular prism. Ahswer the question(s) for your elass, show vour work:
\newline
Af Crases:
\newline
Pe-Algebra:
\newline
he volume of the flgure is g
60
60
60
int, What is the surtace area?
\newline
Bebra:
\newline
co surface area of the three rectangles is S
40
40
40
in?, What is the lume?
\newline
ometry:
\newline
surface area of the figure is
756
i
n
2
756 \mathrm{in}^{2}
756
in
2
, What is the volume?
\newline
20
20
20
.
5
5
5
\newline
Answer:
\newline
Surface Area
Get tutor help
June
232
H
232 \mathrm{H}
232
H
\newline
Overview
\newline
Question Progress
\newline
Exam Paper Progress
\newline
Grade For This Paper
\newline
29
/
80
29 / 80
29/80
Marks
\newline
u
3
3
3
3
3
3
.
\newline
15
15
15
\newline
6
6
6
\newline
T.
\newline
Here is a rectangle and a triangle.
\newline
All measurements are in centimetres.
\newline
The area of the triangle is
12
c
m
2
12 \mathrm{~cm}^{2}
12
cm
2
greater than the area of the rectangle.
\newline
Work out the value of
x
x
x
.
\newline
Optional working
\newline
x
=
x=
x
=
\newline
Answer
\newline
Total marks:
4
4
4
Get tutor help
Narne
\newline
Decompose Fractions into Sums
\newline
Model with Mathematics Look at the list of the names of the months of the year.
\newline
- Draw a visual model to show the fraction of the names of the months that contain the letter
y
y
y
.
\newline
\begin{tabular}{|l|l|}
\newline
\hline Lanuary & July \\
\newline
\hline February & August \\
\newline
\hline March & September \\
\newline
\hline April & October \\
\newline
\hline Mlay & November \\
\newline
\hline June & December \\
\newline
\hline
\newline
\end{tabular}
\newline
- Write an addition equation to model the fraction as a sum of the fractions representing each month.
\qquad
\newline
Write the fraction as a sum of unit fractions. Numerator is
1
1
1
.
\newline
[
2
2
2
]
2
6
=
\frac{2}{6}=
6
2
=
\qquad
[
3
3
3
]
4
5
=
1
5
\frac{4}{5}=\frac{1}{5}
5
4
=
5
1
\newline
(4.)
3
8
=
1
8
\text { (4.) } \frac{3}{8}=\frac{1}{8}
(4.)
8
3
=
8
1
\newline
\qquad
\newline
\qquad
\newline
\qquad
\newline
5
5
5
(
8
8
8
) Model with Mathematics Marlon threw a bowling ball twice and knocked down a total of
6
6
6
pins. Shade the visual fraction models to show two ways he could have knocked down the
6
6
6
pins on the two throws. Then model each with an equation.
\newline
Write the fraction as a sum of two fractions.
\newline
Module
14
14
14
* Lesson
1
1
1
\newline
P
1
1
1
as
Get tutor help
Substitute
s
=
3
s=3
s
=
3
. Area,
A
=
b
h
A=b h
A
=
bh
A
=
3
2
=
9
A=3^{2}=9
A
=
3
2
=
9
Substitute
b
=
5
b=5
b
=
5
and
h
=
2
h=2
h
=
2
.
A
=
5
×
2
=
10
A=5 \times 2=10
A
=
5
×
2
=
10
The area is
9
m
2
9 \mathrm{~m}^{2}
9
m
2
. The area is
10
c
m
2
10 \mathrm{~cm}^{2}
10
cm
2
.
\newline
Check
\newline
1
1
1
. Find the perimeter and area of each figure.
\newline
a)
\newline
−
2.0
k
m
=
2
(
=
Perimeter
=
\begin{array}{r} -2.0 \mathrm{~km}=2( \\ = \\ \text { Perimeter }= \end{array}
−
2.0
km
=
2
(
=
Perimeter
=
\newline
+
+
+
\qquad
\newline
Perimeter
A
=
b
h
A=b h
A
=
bh
0
0
0
\qquad
\newline
b)
\newline
P
=
2
(
b
+
h
)
P=2(b+h)
P
=
2
(
b
+
h
)
\newline
A
=
b
h
=
\begin{aligned} A & =b h \\ & = \end{aligned}
A
=
bh
=
\newline
\qquad
A
=
b
h
A=b h
A
=
bh
3
3
3
\qquad
\newline
=
=
=
\newline
\qquad
\newline
Area
A
=
b
h
A=b h
A
=
bh
0
0
0
\qquad
\newline
Copyright (c)
2006
2006
2006
Pearson Education Canada Inc.
\newline
Area
A
=
b
h
A=b h
A
=
bh
0
0
0
\qquad
Perimeter
A
=
b
h
A=b h
A
=
bh
0
0
0
\qquad
Area
A
=
b
h
A=b h
A
=
bh
0
0
0
\qquad
Get tutor help
MTH
1
1
1
W
1
1
1
\newline
SPIRAL
3
3
3
TAKE HOME ASSIGNMENT
\newline
Communication
\newline
6
6
6
. This is a LEVELLED CHOICE QUESTION. Answer at least one of the following qu
\newline
Create a fully simplified expression for the area of a square with a side length of
3
x
3 x
3
x
.
\newline
area of square
=
(
side length
)
2
or
(
side
)
(
side
)
\text { area of square }=(\text { side length })^{2} \quad \text { or } \quad(\text { side })(\text { side })
area of square
=
(
side length
)
2
or
(
side
)
(
side
)
Get tutor help
19
19
19
The area of a parallelogram
A
B
C
D
\mathrm{ABCD}
ABCD
is
192
s
q
.
c
m
192 \mathrm{sq.} \mathrm{cm}
192
sq.
cm
, and its height is
24
c
m
24 \mathrm{~cm}
24
cm
. Find the length of the base of parallelogram
A
B
C
D
A B C D
A
BC
D
. సమాంతర చతుర్పుజం ABCD యొక్క వైశాల్యం
192
192
192
చ.సెం.మీ. మరియు దాని ఎత్తు
24
24
24
సెం.మీ. అయినా ఆ సమాంతర చతుర్భుజం ABCD యొక్క భూమి పొడవు ను కనుగొనండి.
\newline
A
8
c
m
8 \mathrm{~cm}
8
cm
8
8
8
సెం.మీ.
\newline
B
16
c
m
16 \mathrm{~cm}
16
cm
\newline
16
16
16
సెం.మీ.
\newline
C
168
c
m
168 \mathrm{~cm}
168
cm
\newline
D
216
c
m
216 \mathrm{~cm}
216
cm
\newline
168
168
168
సెం.మీ.
\newline
216
216
216
సెం.మీ.
\newline
20
20
20
Express
16000
16000
16000
in standard form.
\newline
16000
16000
16000
ని ప్రామాణిక రూపంలో వ్యక్తపరచండి.
\newline
A
1.6
×
1
0
4
1.6 \times 10^{4}
1.6
×
1
0
4
\newline
[B]
1.6
×
1
0
3
1.6 \times 10^{3}
1.6
×
1
0
3
\newline
C
192
s
q
.
c
m
192 \mathrm{sq.} \mathrm{cm}
192
sq.
cm
0
0
0
\newline
7
7
7
/
12
12
12
\newline
Section B: Answer the following questions in your ansh.....ets. సెక్షన్ B: మీ జవాబు పత్రంలో ఈ క్రింది ప్రశ్నలకు సమాధానాలు ఇవ్వండి.
\newline
192
s
q
.
c
m
192 \mathrm{sq.} \mathrm{cm}
192
sq.
cm
1
1
1
\newline
21
21
21
a) A student scores
35
35
35
on the first quiz and
50
50
50
on the second quiz. Calculate the percentage increase in their marks. ఒక విద్యార్థికి మొదటి క్విజ్లో స్కోరు
35
35
35
, రెండవ క్విజ్లో స్కోర్
50
50
50
వచ్చిన యెడల, అతనికి వచ్చిన మార్కులలో పెరిగిన శాతమెంత?
\newline
b) A dress is sold at Rs.
92
92
92
with a profit of
192
s
q
.
c
m
192 \mathrm{sq.} \mathrm{cm}
192
sq.
cm
2
2
2
. What is the original cost price of the dress?
\newline
192
s
q
.
c
m
192 \mathrm{sq.} \mathrm{cm}
192
sq.
cm
2
2
2
లాభంతో ఒక డ్రెస్ ను రూ.
92
92
92
కు అమ్మిన, డ్రెస్ యొక్క అసలు ధరను కనుగొనుము.
Get tutor help
7
7
7
\newline
MR
\newline
19
19
19
The area of a parallelogram
A
B
C
D
\mathrm{ABCD}
ABCD
is
192
s
q
.
c
m
192 \mathrm{sq} . \mathrm{cm}
192
sq
.
cm
, and its height
24
c
m
24 \mathrm{~cm}
24
cm
. Find the length of the base of parallelogram
A
B
C
D
A B C D
A
BC
D
. సమాంతర చతుర్తుజం ABCD యొక్క వైశాల్యం
192
192
192
చ.సెం.మీ. మరియు దాశా ఎత్తు
24
24
24
సెం.మీ. అయినా ఆ సమాంతర చతుర్ృుజం ABCD యొక్క భూవి పొడవు ను కనుగొనండి.
\newline
A
8
c
m
8 \mathrm{~cm}
8
cm
\newline
B
16
c
m
16 \mathrm{~cm}
16
cm
\newline
8
8
8
సెం.మీ.
\newline
16
16
16
సెం.మీ.
\newline
C
168
c
m
168 \mathrm{~cm}
168
cm
\newline
D
216
c
m
216 \mathrm{~cm}
216
cm
\newline
168
168
168
సెం.మీ.
\newline
216
216
216
సెం.మీ.
\newline
20
20
20
Express
16000
16000
16000
in standard form.
\newline
16000
16000
16000
ని ప్రామాణిక రూపంలో వ్యక్తపరచండి.
\newline
A
1.6
×
1
0
4
1.6 \times 10^{4}
1.6
×
1
0
4
\newline
B
1.6
×
1
0
3
1.6 \times 10^{3}
1.6
×
1
0
3
\newline
192
s
q
.
c
m
192 \mathrm{sq} . \mathrm{cm}
192
sq
.
cm
0
0
0
\newline
7
7
7
/
12
12
12
\newline
Section B: Answer the following questions in your ansh.....ets. సెక్షన్ B: మీ జవాబు పత్రంలో ఈ క్రింది ప్రశ్నలకు సమాధానాలు ఇవ్వండి.
\newline
192
s
q
.
c
m
192 \mathrm{sq} . \mathrm{cm}
192
sq
.
cm
1
1
1
\newline
21
21
21
a) A student scores
35
35
35
on the first quiz and
50
50
50
on the second quiz. Calculate the percentage increase in their marks. ఒక విద్యార్ధికి మొదటి క్విజ్లో స్కోరు
35
35
35
, రెండవ క్విజ్లో స్కోర్
50
50
50
వచ్చిన యెడల, అతనికి వచ్చిన మార్కులలో పెరిగిన శాతమెంత?
\newline
b) A dress is sold at Rs.
92
92
92
with a profit of
192
s
q
.
c
m
192 \mathrm{sq} . \mathrm{cm}
192
sq
.
cm
2
2
2
. What is the original cost price of the dress?
\newline
192
s
q
.
c
m
192 \mathrm{sq} . \mathrm{cm}
192
sq
.
cm
2
2
2
లాభంతో ఒక డ్రెస్ ను రూ.
92
92
92
కు అమ్మిన, డ్రెస్ యొక్క అసలు ధరను కనుగొనుము.
\newline
[ Turn
0
0
0
Get tutor help
π
=
3.14.1
\pi=3.14 .1
π
=
3.14.1
\newline
(
10
10
10
. Find the area of the shaded region. (Take
π
=
3.14
\pi=3.14
π
=
3.14
.)
\newline
Ans:
\qquad
\newline
As
\newline
15
15
15
. Find the area of the shaded region. (Take
π
=
3.14
\pi=3.14
π
=
3.14
)
Get tutor help
12
12
12
\newline
The illustration shows two circles of radii
4
c
m
4 \mathrm{~cm}
4
cm
and
2
c
m
2 \mathrm{~cm}
2
cm
respectively. The distance between the two centres is
8
c
m
8 \mathrm{~cm}
8
cm
. Find the length of the common tangent
[
A
B
]
[\mathrm{AB}]
[
AB
]
.
\newline
5.29
5.29
5.29
Get tutor help
⇒
5
=
2
x
−
x
=
x
⇒
x
=
5
\Rightarrow 5=2 x-x=x \Rightarrow x=5
⇒
5
=
2
x
−
x
=
x
⇒
x
=
5
\newline
Hence, Shobo's present age
=
5
=5
=
5
years and Shobo's mother's present age
=
6
×
5
=
30
=6 \times 5=30
=
6
×
5
=
30
years
\newline
1
1
1
. A number is such that it is as much greater than
84
84
84
as it is less than
108
108
108
. Find it.
Get tutor help
Question
3
3
3
\newline
1
p
t
s
1 \mathrm{pts}
1
pts
\newline
You wish to test the following claim
(
H
a
)
\left(\mathrm{H}_{\mathrm{a}}\right)
(
H
a
)
at a significance level of
α
=
0.05
\alpha=0.05
α
=
0.05
.
\newline
H
o
:
p
=
0.42
H
a
:
p
>
0.42
\begin{array}{l} \mathrm{H}_{\mathrm{o}}: p=0.42 \\ \mathrm{H}_{\mathrm{a}}: p>0.42 \end{array}
H
o
:
p
=
0.42
H
a
:
p
>
0.42
\newline
You obtain a sample of size
n
=
143
n=143
n
=
143
in which there are
73
73
73
successful observations. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.
\newline
What is the test statistic? Round to
2
2
2
decimal places.
\newline
2
2
2
.
192
192
192
\newline
2
2
2
.
19
19
19
Get tutor help
The diagram shows a container consisting of a square open top with rectangular sides, each
10
x
c
m
10 x \mathrm{~cm}
10
x
cm
by
h
c
m
h \mathrm{~cm}
h
cm
, and an inverted regular pyramid. The perpendicular height of the pyramid is
10
r
c
m
12
x
c
m
10 \mathrm{r} \mathrm{cm} 12 x \mathrm{~cm}
10
r
cm
12
x
cm
.
\newline
(i) Find an expression in terms of
x
x
x
for the area of one triangular face of the pyramid.
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You wish to test the following claim
(
H
a
)
\left(H_{a}\right)
(
H
a
)
at a significance level of
α
=
0.10
\alpha=0.10
α
=
0.10
.
\newline
H
o
:
μ
=
60.8
H
a
:
μ
>
60.8
\begin{array}{l} H_{o}: \mu=60.8 \\ H_{a}: \mu>60.8 \end{array}
H
o
:
μ
=
60.8
H
a
:
μ
>
60.8
\newline
You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:
\newline
\begin{tabular}{|r|}
\newline
\hline data \\
\newline
\hline
64
64
64
.
8
8
8
\\
\newline
\hline
33
33
33
.
6
6
6
\\
\newline
\hline
69
69
69
.
6
6
6
\\
\newline
\hline
54
54
54
.
7
7
7
\\
\newline
\hline
79
79
79
\\
\newline
\hline
69
69
69
.
6
6
6
\\
\newline
\hline
86
86
86
.
9
9
9
\\
\newline
\hline
69
69
69
.
6
6
6
\\
\newline
\hline
93
93
93
.
5
5
5
\\
\newline
\hline
99
99
99
.
2
2
2
\\
\newline
\hline
90
90
90
.
6
6
6
\\
\newline
\hline
76
76
76
.
7
7
7
\\
\newline
\hline
65
65
65
.
8
8
8
\\
\newline
\hline
\newline
\end{tabular}
\newline
What is the critical value for this test? (Report answer accurate to three decimal places.)
\newline
critical value
=
1.356
=1.356
=
1.356
\newline
What is the test statistic for this sample? (Report answer accurate to three decimal places.)
\newline
test statistic
=
=
=
□
\square
□
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8
8
8
: Circular Measure
\newline
In the diagram,
O
O
O
is the centre of the circle
A
C
B
Q
A C B Q
A
CBQ
of radius
20
c
m
.
A
P
B
20 \mathrm{~cm} . A P B
20
cm
.
A
PB
jom arc of a circle with centre
C
C
C
and angle
A
C
B
=
1.4
A C B=1.4
A
CB
=
1.4
radians.
C
P
C P
CP
bisects
∠
A
C
B
\angle A C B
∠
A
CB
.
\newline
(a) Find otuge angle
A
O
B
A O B
A
OB
in radians.
\newline
[
1
]
[1]
[
1
]
\newline
(b) Show that
A
C
=
30.59
A C=30.59
A
C
=
30.59
cm correct to
2
2
2
decimal places.
\newline
(c) Calculate the perimeter of the shaded region.
\newline
(d) Calculate the area of the shaded region.
\newline
[Ans: (a)
A
C
B
Q
A C B Q
A
CBQ
0
0
0
(b)
30
30
30
.
59
59
59
\newline
(c)
A
C
B
Q
A C B Q
A
CBQ
1
1
1
\newline
(d)
A
C
B
Q
A C B Q
A
CBQ
2
2
2
]
\newline
[
4
4
4
]
\newline
[
3
3
3
]
\newline
[
3
3
3
]
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Graph a right triangle with the two points forming the hypotenuse. Using the sides, find the distance between the two points, to the nearest tenth (if necessary).
\newline
(
7
,
−
7
)
and
(
1
,
1
)
(7,-7) \text { and }(1,1)
(
7
,
−
7
)
and
(
1
,
1
)
\newline
*Click twice to draw a line. Click a segment to erase it.
∗
{ }^{*}
∗
\newline
Answer Attempt
1
1
1
out of
2
2
2
\newline
Leg
1
1
1
:
□
\square
□
Leg
2
2
2
:
□
\square
□
Distance:
□
\square
□
\newline
Submit Answer
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7
7
7
A logo is designed with four identical squares. A triangle is drawn in each square as shown. Find the total area of the four triangles.
\newline
Area of figure
=
8
c
m
×
5
c
m
=
64
c
m
2
\begin{aligned} \text { Area of figure } & =8 \mathrm{~cm} \times 5 \mathrm{~cm} \\ & =64 \mathrm{~cm}^{2}\end{aligned}
Area of figure
=
8
cm
×
5
cm
=
64
cm
2
\newline
Ans:
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1
1
1
For a square of sides
x
c
m
x \mathrm{~cm}
x
cm
, write the formulae for its perimeter
P
P
P
and area
A
A
A
in terms of
x
x
x
.
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The scatter plot and line of best fit below show the length of
12
12
12
people's femur (the long leg bone in the thigh) and their height in centimeters. Based on the line of best fit, what would
b
b
b
the predicted height for someone with a femur length of
55
c
m
55 \mathrm{~cm}
55
cm
?
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A
(
x
)
=
(
8
+
2
x
)
2
A(x) = (8 + 2x)^2
A
(
x
)
=
(
8
+
2
x
)
2
\newline
Carmen wants to add a border around a square picture. The function shows the area of the picture in square inches if it has a border
x
x
x
inches wide. What are the dimensions of the picture without the border?
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Extend
\newline
14
14
14
. From the top of a
35
−
m
35-\mathrm{m}
35
−
m
-tall building, an observer sees a truck heading toward the building at an angle of depression of
1
0
∘
10^{\circ}
1
0
∘
. Ten seconds later, the angle of depression to the truck is
2
5
∘
25^{\circ}
2
5
∘
.
\newline
a) Determine the distance that the truck has travelled. Express your answer to the nearest metre.
\newline
b) If the speed limit for the area is
40
k
m
/
h
40 \mathrm{~km} / \mathrm{h}
40
km
/
h
, is the truck driver following the speed limit? Explain.
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QUESTION
1
1
1
\newline
Metsis are bolated from ther ores by reacting the metal exide with carbon as shown in the following equation
\newline
z
M
O
(
s
)
+
C
(
s
)
→
Z
M
(
s
)
+
C
O
2
(
g
)
zMO(s) + C(s) \rightarrow ZM(s) + CO_{2}(g)
z
MO
(
s
)
+
C
(
s
)
→
ZM
(
s
)
+
C
O
2
(
g
)
\newline
18.6
g
18.6g
18.6
g
of a pure metal oxibe reacted with excess carbon
\newline
4.98
L
4.98L
4.98
L
of
\newline
C
O
2
CO_{2}
C
O
2
gas was formed at
\newline
20
0
∘
C
200^{\circ}C
20
0
∘
C
and the pressure at
\newline
1.80
a
t
m
1.80atm
1.80
a
t
m
. What is the ibentty of metal, M?
\newline
Hg
\newline
Ng
\newline
Cu
\newline
Ca
\newline
Un
\newline
QUESTION
2
2
2
\newline
86.8
g
86.8g
86.8
g
\newline
57.9
g
57.9g
57.9
g
\newline
2359
g
2359g
2359
g
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QUESTION
1
1
1
\newline
Metals are bolated frem ther ores by reacting the metal exide with carbon as shown in the following equation.
\newline
Z
M
O
(
s
)
+
C
(
s
)
→
Z
M
(
s
)
+
C
O
2
(
g
)
\mathrm{ZMO}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{ZM}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})
ZMO
(
s
)
+
C
(
s
)
→
ZM
(
s
)
+
CO
2
(
g
)
\newline
II
18.6
g
18.6 \mathrm{~g}
18.6
g
of a pure metal oxibe reacted with excess cartion
4.98
L
4.98 \mathrm{~L}
4.98
L
of
C
O
2
\mathrm{CO}_{2}
CO
2
gas was formed at
20
0
∘
C
200^{\circ} \mathrm{C}
20
0
∘
C
and the pressure at
1.80
a
t
m
1.80 \mathrm{~atm}
1.80
atm
. What is the ibentity of metal,
M
\mathrm{M}
M
?
\newline
H
g
\mathrm{Hg}
Hg
\newline
N
g
9
\mathrm{Ng}_{9}
Ng
9
\newline
C
u
\mathrm{Cu}
Cu
\newline
C
a
\mathrm{Ca}
Ca
\newline
Un
\newline
4.98
L
4.98 \mathrm{~L}
4.98
L
0
0
0
\newline
4.98
L
4.98 \mathrm{~L}
4.98
L
0
0
0
\newline
QUESTION
2
2
2
\newline
4.98
L
4.98 \mathrm{~L}
4.98
L
2
2
2
\newline
4.98
L
4.98 \mathrm{~L}
4.98
L
3
3
3
\newline
4.98
L
4.98 \mathrm{~L}
4.98
L
4
4
4
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rimeter, Area, and Voly.me
\newline
ce area of a trizhgular prism
\newline
surface area of this triangular prism. Be sure to include the correct unit in your answer.
\newline
□
\square
□
\newline
□
\square
□
c
m
\mathrm{cm}
cm
\newline
xplanation
\newline
Check
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